Question

In the figure given below, P and Q are centres of two circles, intersecting at B and C, and ACD is a straight line.

If ∠APB = 150° and ∠BQD = x°, find the value of x.

Answer 1

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc AEB subtends ∠APB at the centre and ∠ACB at C on the circle.

∴ ∠APB = 2∠ACB
⇒ $\angle ACB=\frac{150°}{2}=75°$ ...(1)

Since ACD is a straight line, ∠ACB + ∠BCD = 180^∘
⇒ ∠BCD = 180^∘ − 75^∘
⇒ ∠BCD = 105^∘ ...(2)

Also, arc BFD subtends reflex ∠BQD at the centre and ∠BCD at C on the circle.

∴ reflex ∠BQD = 2∠BCD
⇒ $\mathrm{reflex}\angle BQD=2\left(105°\right)=210°$ ...(3)

Now,
reflex ∠BQD + ∠BQD = 360^∘
⇒ 210^∘ + x = 360^∘
⇒ x = 360^∘ − 210^∘
⇒ x = 150^∘

Hence, x = 150^∘.